Using the BQD Iterator it's possible to show a connection between squares and integers raised to an arbitrarily high power.

The appropriate BQD iterator is, given:

The appropriate BQD iterator is, given:

u

^{2}+ (m - 1)v

^{2}= F

then it must also be true that

(u - (m - 1)v)

^{2}+ (m - 1)(u + v)

^{2}= m*F

So if you start the iterator with u = v = 1, or u = v = -1, F = m.

That means that in general there must always exist nonzero x and y, such that for an integer n equal to 0 or higher, and an integer m equal to 3 or higher:

**x**

^{2}+ (m-1)y^{2}= m^{n+1}I have m raised to n+1 so that n is a count of iterations, if you want m raised to the nth power you just start n at 1 instead of starting at 0, which I prefer.

Wow. That's really cool. Let's try it.

Let's let m = 5, and u = v = 1. Then, iterator is:

(u - 4v)

^{2}+ 4(u + v)

^{2}= 5*F

Start is:

1

^{2}+ 4*1

^{2}= 5

then it must also be true that

(-3)

^{2}+ 4(2)

^{2}= 25 = 5

^{2}

Next iteration: (-11)

^{2}+ 4(-1)

^{2}= 125 = 5

^{3}

And third iteration: (-7)

^{2}+ 4(-12)

^{2}= 625 = 5

^{4}

Fourth iteration: (41)

^{2}+ 4(-19)

^{2}= 3125 = 5

^{5}

Fifth iteration: (117)

^{2}+ 4(22)

^{2}= 15625 = 5

^{6}

Sixth iteration: (29)

^{2}+ 4(139)

^{2}= 78125 = 5

^{7}

So it's easy to see how in general:

**x**

^{2}+ 4y^{2}= 5^{n+1}And calculating x and y is easy with the iterator. I'll note that it's important to keep up with signs. But of course x or y can be positive or negative, which means you can switch signs for either at each iteration.

That's how you can figure out each x and y that will work. But the math can also simply force in a factor of 5 into each, which means 25 divides off, pushing you backwards.

Oh yeah, and with m-1 a square you end up with a sum of squares, which is what happened with my example above, which is the first one since m must be 3 or greater. So if m - 1 is a square, then m

^{n}can be written as a sum of exactly two squares for any n greater than 0. That's cool too. Wow, love these infinity results.

So it is proven that in general squares have a connection to integers raised to an arbitrarily high power.

Actually, to be more specific, it's proven that m squares where m-1 of them are the same, can always be found to sum to m raised to an arbitrarily high power.

Watching those numbers behave as predicted is a thrill unlike any other. That thrill is what you want to share.

That's pretty cool. I wonder, did anyone know that before my research? Will have to go do some web searching.

James Harris